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Homework Help: Prove that a limit does not exist

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that limx→2 (1/(x – 2)) does not exist.

    2. The attempt at a solution

    My approach to this problem is to use the sequential criterion for nonexistence, that is, if {x_n} and {y_n} both converge to a (2 in this case), but limx→a f(x_n) is not equal to limx→a f(y_n), then limx→a f(x) does not exist (a being 2 in this case).

    I was able to find two sequences, {x_n} and {y_n}, both converging to 2 (I chose {x_n} = 1.5 + (1/n) and {y_n} = 2.5 – (1/n) as converging sequences). But when I try to plug these values in f(x_n) and f(y_n), I get (2 – n) or an integral multiple of it in the denominator which doesn’t help me when taking the limit for either f(x_n) or f(y_n) when n→2.

    I also tried some other combinations but I seem to run into the same problem, division by zero when taking the limit for either f(x_n) or f(y_n), can you please tell me if I am missing something or if I am doing something wrong here? Thanks in advance for all your help.
  2. jcsd
  3. Oct 29, 2011 #2


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    You don't want n → 2. You need {xn} → 2 as n → ∞, and {yn} → 2 as n → ∞, while f(x_n) ≠ f(y_n) as n → ∞.
  4. Oct 29, 2011 #3


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    What you have done (n->infinity, [itex]x_n[/itex]-> 2) is perfectly valid (I think Sammy S misunderstood what you were saying) but, since you get [itex]x_n-2[/itex] in the denominator, the denominator is going to 0 while the numerator is always 1. Of course, that does not exist- it goes to infinity.

    (Note: saying [itex]lim_{x\to 2} f(x)= \infty[/itex] is just another way of say the limit does not exist. "Infinity" is NOT a number so is not a limit.)

    Or, try [itex]x_n= 2+ 1/n[/itex], [itex]y_n= 2- 1/n[/itex]. As n goes to infinity, [itex]x_n[/itex] goes to 2 "from above" and the limit is +infinity while [itex]y_n[/itex] goes to 2 "from below" and the limit is -infinity.
  5. Oct 29, 2011 #4


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    it is possible to use the definition of the limit directly, here.

    suppose that f DID have a limit at 2, call it L.

    could this limit be 0? no, because if we choose, say ε = 1,

    then anytime |x-2| < 1, |f(x)| = 1/|x-2| > 1, so we can't find a δ.

    so if L exists, L ≠ 0. therefore |L| > 0, and we can use that as our ε.

    but if 0 < |x-2| < 1/(2|L|),

    then |1/(x-2) - L| ≥ 1/|x-2| - |L|...can you see how to finish?

    no matter how small a δ we choose, there will be some x for which |x-2| < 1/(2|L|),

    which means....?
  6. Oct 30, 2011 #5
    Thanks for the answers everyone; here is what I have done so far:

    Choose {x_n} = 2 + (1/n). Then limn→∞ (x_n) = 2.
    Hence 1/((x_n) – 2) = n = f(x_n), and limn→∞ f(x_n) = limn→∞ n = ∞

    Now choose {y_n} = 2 - (1/n). Then limn→∞ (y_n) = 2.
    Hence 1/((y_n) – 2) = (-n) = f(y_n), and limn→∞ f(y_n) = limn→∞ (-n) = -∞

    Since limn→∞ f(x_n) does not equal limn→∞ f(y_n), then limn→∞ f(x) does not exist.

    Thus limx→2 (1/(x – 2)) does not exist.
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